std::hash specialization using sfinae?

问题:

As an exercise I was trying to see if I could use SFINAE to create a std::hash specialization for std::pair and std::tuple when all of its template parameters are of an unsigned type. I have a little experience with them, but from what I understand the hash function needs to have already been templated with a typename Enabled = void for me to add a specialization. I'm not really sure where to go from here. Here's an attempt which doesn't work.

#include <functional>
#include <type_traits>
#include <unordered_set>
#include <utility>

namespace std {
template <typename T, typename Enabled = void>
struct hash<std::pair<T, T>, std::enable_if_t<std::is_unsigned<T>::value>>
{
    size_t operator()(const std::pair<T, T>& x) const
    {
        return x;
    }
};
}; // namespace std


int
main(int argc, char ** argv)
{
    std::unordered_set<std::pair<unsigned, unsigned>> test{};
    return 0;
}

Error:

hash_sfinae.cpp:7:42: error: default template argument in a class template partial specialization
template <typename T, typename Enabled = void>
                              ^
hash_sfinae.cpp:8:8: error: too many template arguments for class template 'hash'
struct hash<std::pair<T, T>, std::enable_if_t<std::is_unsigned<T>::value>>

It's about what I expected because I'm trying to extend the template parameters to hash... But I'm not sure the technique to handle these cases then. Can someone help me understand?

回答1:

You are not supposed to specialize std::hash for types that doesn't depend on a type you defined yourself.

That said, this hack might work:

template<class T, class E>
using first = T;

template <typename T>
struct hash<first<std::pair<T, T>, std::enable_if_t<std::is_unsigned<T>::value>>>
{
    size_t operator()(const std::pair<T, T>& x) const
    {
        return x;
    }
};

Really, though, don't do this. Write your own hasher.